Let \(D\subseteq\CC\) be a connected open set, let \(f\) be holomorphic on \(D\) and not identically zero, and let \(z_0\in D\) with \(f(z_0)=0\). Then there exist an integer \(k\ge1\) and a function \(g\) holomorphic on \(D\) with \(g(z_0)\neq0\) such that \[ f(z)=(z-z_0)^kg(z)\quad\text{for all } z\in D. \]
Choose \(r>0\) with \(D(z_0,r)\subseteq D\). Since \(f\) is not identically zero on the connected set \(D\), not every derivative \(f^{(j)}(z_0)\) vanishes, and \(f(z_0)=0\); let \(k\ge1\) be the least integer with \(f^{(k)}(z_0)\neq0\). On \(D(z_0,r)\), \[ f(z)=\sum_{j\ge k}\frac{f^{(j)}(z_0)}{j!}(z-z_0)^j=(z-z_0)^k\sum_{j\ge0}\frac{f^{(j+k)}(z_0)}{(j+k)!}(z-z_0)^j. \] Define \[ g(z)\coloneqq\begin{cases}\displaystyle\sum_{j\ge0}\frac{f^{(j+k)}(z_0)}{(j+k)!}(z-z_0)^j, & z\in D(z_0,r),\\[1ex] \dfrac{f(z)}{(z-z_0)^k}, & z\in D\setminus\theset{z_0}.\end{cases} \] The two formulas agree on \(D(z_0,r)\setminus\theset{z_0}\), so \(g\) is well defined; it is holomorphic on \(D(z_0,r)\) as a convergent power series and on \(D\setminus\theset{z_0}\) as a quotient of holomorphic functions, hence on \(D\). Moreover \(g(z_0)=f^{(k)}(z_0)/k!\neq0\) and \(f(z)=(z-z_0)^kg(z)\) on \(D\).